A cube of side 2 m is placed in front of a large concave mirror of focal length 1 m is such a way that the face A of the cube is at a distance of 3 m and the face B at a distance of 5 m form the mirror. (i) Calculate the distance between the images of the faces A and B. (ii) Determine the height of the images of the faces A and B. (iii) Will the image of the cube be a cube?

Explanation

Distance of the face A from the mirror, `u_(1)=-3m`
[Fig.1.46], focal length of the mirror , f = - 1 m

Let the image distance of the face A from the mirror be `v_(1) m .`
`"We know", " " (1)/(v)+(1)/(u)=(1)/(f) or, (1)/(v_(1))-(1)/(3)=-(1)/(1) or, v_(1)=-1.5 m `
Distance of the face B from the mirror `u_(2)=-5m,` the image distance of this face = `v_(2) m`
`"We know", (1)/(v)+(1)/(u)=(1)/(f)`
`or, " " (1)/(v_(2))-(1)/(5)=-(1)/(1)`
`or, " " v_(2)=-1.25m`
So, the distance between the image of the faces A and B `=v_(1)-v_(2)=1.5-1.25=0.25m`
[taking magnitude of `v_(1) and v_(2)`]
`(ii) " "` Magnification in the first case,
`m_(1)=-(v_(1))/(u_(1))=-(-1.5)/(-3)=-0.5`
`therefore " " ("height of the image of the face" A(I_(A)))/("height of the face"A(O_(A)))=-0.5`
`or, " " I_(A) =-0.5xx2=-1m`
Againing, magnification in the second case,
`m_(2)=(-v_(2))/(u_(2))=(-1.25)/(=5)=-0.25`
`therefore` Height of the image of the face B,
`I_(B)=m_(2)xx"height of the face B"`
`=-0.25 xx 2 = - 0.5 m`
`(iii) " "` So, it is seen that the height of the image of the face A and that of the face B are not equal. So, the image of the cube will no more be a cube.